I was investigating rigid body motion in more than 3 spatial dimensions, and it yielded some very cool results. Just letting go and finally treating bivectors as bivectors is simply a wonderful experience. The moment of inertia tensor is ugly and stupid, and I'm glad that there's a prettier way to calculate these things.

Living in a three-dimensional world is sad because we have all these bad habits when it comes to thinking about rotations and stuff. Four-dimensional rotations are so cool.

Eebster the Great wrote:And every non-discrete, nontrivial topological space also has subsets which are neither closed nor open.

I guess it depends on what you mean by non-trivial, but this doesn't seem right. Take the space on two points, A and B, whose open sets are {}, {A}, and {A, B}. More generally, take the open sets to be an ultrafilter on the underlying space, and then toss in the empty set.

(∫|p|2)(∫|q|2) ≥ (∫|pq|)2Thanks, skeptical scientist, for knowing symbols and giving them to me.

Oh, of course, thanks! I was trying to work out whether there was something like that but got stuck in circles that kept coming back to "well obviously, that's because sqrt(x) is the inverse of x^2".EDIT: Which was not helping me make the link.

Given 3D eucledian space with n line segments of unit length that adjoin perpendicularly, can you make a loop? (like making a chain of 6-sided sodium atoms and rods with a molecular model set)

n<3 doesn't count; n=3 doesn't work because you're always √2 away from the starting point after two segments (and the plane of possible third segments doesn't intersect the starting point).For n=4 this is simply a square. And for every larger even number n=2k>4 you can just pick two opposing sides of the square and insert line segments in between.But does it also work for odd numbers n=2k+1>4? My gut feeling says no for at least n=5, but I can't prove it (at this time).

Flumble wrote:But does it also work for odd numbers n=2k+1>4? My gut feeling says no for at least n=5, but I can't prove it (at this time).

Also, does adding dimensions have any effect?

Your gut feeling is correct, and adding more dimensions doesn't help.

For the 3D case, the segments are unit vectors in the X, Y, and Z directions, and the oppositely-directed unit vectors -X, -Y, -Z. Because X, Y & Z are mutually perpendicular no combination of {X, -X, Y, -Y} sums to a scalar multiple of Z; similarly, you can't build a scalar multiple of Y just using scalar multiples of Xs & Zs, or a scalar multiple of X just using Ys & Zs. So if a path built from elements of {X, -X, Y, -Y, Z, -Z} returns to its starting point the number of X segments must be equal to the number of -X segments, and the same goes for Y & -Y and Z & -Z. So the total number of X & -X segments must be even, similarly for the Y & -Y and Z & -Z. And so the total number of all segments must be even.

PM 2Ring wrote:For the 3D case, the segments are unit vectors in the X, Y, and Z directions, and the oppositely-directed unit vectors -X, -Y, -Z.

I think the OP is asking for the more general case where the segments are not necessarily axis-aligned.

indeed (wait, am I "OP"? after 9 pages?)

Thanks for the n>=7 case, jaap. (I was doubtful whether n=7 even has enough degrees of freedom, or indeed whether any odd number were possible, as making it fit in practice/head doesn't entail that it has right angles in theory)

Non-adjacent vertices must be a distance of sqrt(2) apart. So what you have is essentially this pentagon:

right-pentagon.png (3.7 KiB) Viewed 5404 times

The three triangles in this pentagon all have known side lengths, so the only degrees of freedom left are the angle between the triangles, i.e. how far to fold along the two thin lines of the diagram. You can fold ABC up/down until angle C is 90 degrees, and similarly with triangle ADE to make D a right angle. You now have no freedom any more to make A a right angle, and it really would be very lucky if it were. If you fold both triangles upwards, or both downwards, then BCDE will lie in a plane, distance BE will be 1, and so A will be 60 degrees. I have not (yet) checked what angle A becomes when you fold the triangles in opposite directions, and it does look like it could be close to 90 degrees. (Edit: it's 110.9 degrees if I did it right)

Thanks jaap, that proof is both easy to understand and rigorous (...enough to me; maybe a math professor will ask for an argument why rotating B over AC only has two angles for which BC⟂CD, but we know it's true).

Now dare I ask: is it possible to make a 5-chain in 4D or higher? (I promise I won't ask about properties of these constructions in non-Eucledian space )

Flumble wrote:Now dare I ask: is it possible to make a 5-chain in 4D or higher? (I promise I won't ask about properties of these constructions in non-Eucledian space :P )

Yes. Start with the same pentagon, with the two triangles folded in opposite directions, one in the positive z direction the other in the negative z direction. Now four of the angles are 90 degrees, only the angle at A is larger than that. You can fold the two triangles into the 4th dimension, to reduce the z-coordinates enough to make A a right angle.

Flumble wrote:(I promise I won't ask about properties of these constructions in non-Eucledian space )

Oh, in hyperbolic space it's a piece of cake. Remember that in non-Euclidean geometries the angle sum of a triangle isn't constant but instead is a linear function of the area. In hyperbolic space the sum is always less than pi; you can make the angle as small as you like by making the triangle large enough. Extending this to pentagons, we can therefore make a regular pentagon in hyperbolic space with each vertex angle being a right angle.

Here's a Poincaré disk representation of a hyperbolic tessellation of congruent regular pentagons, from Wikipedia.The angles in these pentagons are all right angles since 4 pentagons meet at a vertex.

Is there a name for the curve defined by the trajectory of a projectile experiencing gravity and quadratic drag? I was trying to come up with a concise way to describe the difference in the paths a fastball and curveball take. The former doesn't follow a parabolic trajectory of course, due to drag, but it still drops less than a curveball.

I don't know any, but I can tell you that wikipedia states that for low velocities (actually, low Reynolds numbers) you can approximate the trajectory with linear drag (so you can solve things analytically):